Page 24 - Lab Manual & Project class 12

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

(i) What is meant by the term, calorimeter constant?

(ii) Why is ∆ H for some substances negative while for others it is positive?
Sol
(iii) How does ∆ H vary with temperature?
Sol
(iv) Will the enthalpy change for dissolution of same amount of anhydrous copper sulphate and
hydrated copper sulphate in the same amount of water be the same or different? Explain.

(v) How will the solubility of copper sulphate and potassium nitrate be affected on increasing
Maxbrain Chemistry
the temperature? Explain.





To determine the enthalpy of neutralisation of a strong acid (HCl)
with a strong base (NaOH).



+
A neutralisation reaction involves the combination of H (aq) ions
–
furnished by an acid and OH (aq) ions furnished by a base,
evidently leading to the formation of H O (l). Since the reaction
2
envisages bond formation, therefore, this reaction is always
exothermic. Enthalpy of neutralisation is defined as the amount
+
of heat liberated when 1mol of H ions furnished by acid combine
–
with 1 mole of OH ions furnished by base to form water. Thus:
+ –
H (aq) + OH (aq) → H O (l), ∆ H is negative
2 neut
(Acid) (Base)

where ∆ H is known as enthalpy of neutralisation.
neut
If both the acid and the base are strong then for the
formation of 1 mol H O (l), always a fixed amount of heat, viz,
2
–1
57 kJ mol is liberated. If any one of the acid or the base is
weak or if both of these are weak, then some of the heat
liberated is used for the ioni sation of the acid or base or both
of them (as the case may be) and the amount of heat liberated
–1
is less than 57 kJ mol .



24-04-2018